Question #2997a

1 Answer
May 28, 2016

Answer:

#"2.5 g sucrose"#

Explanation:

Assuming that you're dealing with percent concentration by mass, #"% m/m"#, you can determine the mass of sucrose present in #"1.0 kg"# of that solution by using its concentration as a conversion factor.

This is essentially what a solution's concentration does -- it helps you go from, in this case, mass of solution to mass of solute or vice versa.

A #0.25%# by mass solution will contain #"0.25 g"# of sucrose, your solute, for every #"100 g"# of solution. So, if every #"100 g"# of this solution will contain #"0.25 g"# of sucrose, it follows that #"1.0 kg"# will contain

#1.0 color(red)(cancel(color(black)("kg"))) * (10^3color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * overbrace("0.25 g sucrose"/(100color(red)(cancel(color(black)("g solution")))))^(color(purple)("= 0.25% m/m")) = color(green)(|bar(ul(color(white)(a/a)color(black)("2.5 g sucrose")color(white)(a/a)|)))#